BPSC 47th Exam Prelims 2005 Solved Paper

Question : 123 of 150

Marks: +1, -0

A cyclist covers the distances of

15 {km}, 20 {km}

and

25 {km}

with respective speeds

15 {km} \/ {h}, 25

{km} \/ {h}

and

15 {km} \/ {h}

. His average speed was

Solution:

Time in covering first

15 {km}

=\;{15}/{15}=1

hour.
Time taken in covering next

20 {km}

=\;{20}/{25}=\;{4}/{5}

hour.
Time taken in covering last

25 {km}

=\;{25}/{15}=\;{5}/{3}

hour.

∵

Average speed

=\;{\;\text " Total Distance "\;}/{\;\text " Total Time "\;}

=\;{15+20+25}/{1+\;{4}/{5}+\;{5}/{3}}

=\;{60}/{{42}/{15}}=21.40 {km} {/} {h}

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